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3 years ago

Do you think it is appropriate to have diet pop available in your school?

Chemistry

1 answer:

pashok25 [27]3 years ago

3 0

Answer:

I think so

Explanation:

It would provide an extra energy boost with lower sugars. Students bring drinks to school anyways so it would be nice to offer some that aren't as detrimental.

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A sample of nitrogen gas has the temperature drop from 250.°c to 150.°c at constant pressure. what is the final volume if the in

Leviafan [203]

Charles law gives the relationship between temperature of gas and volume of gas.
It states that for a fixed amount of gas, temperature is directly proportional to volume of gas.
V / T = k
where V- volume , T - temperature and k - constant
$\frac{V1}{T1} = \frac{V2}{T2}$
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation.
T1 = 250 °C + 273 = 523 K
T2 = 150 °C + 273 = 423 K
Substituting the values in the equation,

$\frac{310 mL}{523 K} = \frac{V}{423K}$
V = 251 mL
the new volume is 251 mL

6 0

3 years ago

How many moles of nitrogen, N, are in 61.0 g of nitrous oxide, N2O?

Oksana_A [137]

Answer:

2.77 mol N

Explanation:

M(N2O) = 2*14 + 16 = 44 g/mol

61.0 g * 1 mol/44g = (61/44) mol N2O

N2O ---- 2N

1 mol 2 mol

(61/44) mol x mol

x = (61/44)*2/1 = 2.77 mol N

4 0

3 years ago

Bromine reacts with nitric oxide to form nitrosyl bromide as shown in this reaction: br2(g) + 2 no(g) → 2 nobr(g) a possible mec

Svetach [21]

The overall reaction is given by:

$Br_{2}(g) + 2 NO(g) \rightarrow 2 NOBr(g)$

The fast step reaction is given as:

$NO(g) + Br_{2}(g) \rightleftharpoons NOBr_{2}(g) (fast; k_{eq}= \frac{k_{1}}{k_{-1}})$

The slow step reaction is given as:

$NOBr_{2}(g) + NO(g) \rightarrow 2 NOBr(g)$ (slow step $k_{2}$ )

Now, the expression for the rate of reaction of fast reaction is:

$r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]$

The expression for the rate of reaction of slow reaction is:

$r_{2}=k_{2}[NOBr_{2}] [NO]$

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as $[NOBr_{2}]$ takes place in this reaction.

The expression of rate of formation is:

$\frac{d(NOBr)}{dt}=r_{2}$

= $k_{2}[NOBr_{2}][NO]$ (1)

Now, consider that the fast step is always is in equilibrium. Therefore, $r_{1}=0$

$k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]$

$[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]$

Substitute the value of $[NOBr_{2}]$ in equation (1), we get:

$\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]$

= $k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]$

= $\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]$

Thus, rate law of formation of $NOBr$ in terms of reactants is given by $\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]$ .

4 0

3 years ago

How did Neils Bohr change the model of the atom

kipiarov [429]

Bohr changed the model of the atom by proposing that electrons travelled in circular orbits with specific energy levels.

6 0

2 years ago

CAN SOMEONE ANSWER ME!! 5-6

docker41 [41]

Answer:

As the axial tilt increases, then the seasonal contrast increases so that winters are colder and summers are warmer in both hemispheres. The northern hemisphere is tipped away from the Sun, producing short days and a low sun angle. What kind of effect does the earth's tilt and subsequent seasons have on our length of daylight (defined as sunrise to sunset). Over the equator, the answer is not much.

7 0

2 years ago